3.1.39 \(\int \frac {x^2 (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^2} \, dx\) [39]

Optimal. Leaf size=144 \[ -\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x (a+b \text {ArcSin}(c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {i (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{c^3 d^2}-\frac {i b \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{2 c^3 d^2}+\frac {i b \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{2 c^3 d^2} \]

[Out]

1/2*x*(a+b*arcsin(c*x))/c^2/d^2/(-c^2*x^2+1)+I*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c^3/d^2-1/2*
I*b*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d^2+1/2*I*b*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d^2-1
/2*b/c^3/d^2/(-c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4791, 4749, 4266, 2317, 2438, 267} \begin {gather*} \frac {i \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^3 d^2}+\frac {x (a+b \text {ArcSin}(c x))}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {i b \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{2 c^3 d^2}+\frac {i b \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{2 c^3 d^2}-\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^2,x]

[Out]

-1/2*b/(c^3*d^2*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x]))/(2*c^2*d^2*(1 - c^2*x^2)) + (I*(a + b*ArcSin[c*x]
)*ArcTan[E^(I*ArcSin[c*x])])/(c^3*d^2) - ((I/2)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^3*d^2) + ((I/2)*b*Pol
yLog[2, I*E^(I*ArcSin[c*x])])/(c^3*d^2)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4749

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sec[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4791

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[f
*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + (-Dist[f^2*((m - 1)/(2*e*(p + 1
))), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*f*(n/(2*c*(p + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^2} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {b \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 c d^2}-\frac {\int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{2 c^2 d}\\ &=-\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac {\text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 d^2}\\ &=-\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}+\frac {b \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 d^2}-\frac {b \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 d^2}\\ &=-\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac {(i b) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {(i b) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}\\ &=-\frac {b}{2 c^3 d^2 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{2 c^2 d^2 \left (1-c^2 x^2\right )}+\frac {i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{c^3 d^2}-\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}+\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{2 c^3 d^2}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(463\) vs. \(2(144)=288\).
time = 0.11, size = 463, normalized size = 3.22 \begin {gather*} -\frac {a x}{2 c^2 d^2 \left (-1+c^2 x^2\right )}+\frac {a \log (1-c x)}{4 c^3 d^2}-\frac {a \log (1+c x)}{4 c^3 d^2}+\frac {b \left (\frac {\sqrt {1-c^2 x^2}-\text {ArcSin}(c x)}{4 c^3 (-1+c x)}-\frac {\sqrt {1-c^2 x^2}+\text {ArcSin}(c x)}{4 c^2 \left (c+c^2 x\right )}+\frac {\frac {3 i \pi \text {ArcSin}(c x)}{2 c}-\frac {i \text {ArcSin}(c x)^2}{2 c}+\frac {2 \pi \log \left (1+e^{-i \text {ArcSin}(c x)}\right )}{c}-\frac {\pi \log \left (1+i e^{i \text {ArcSin}(c x)}\right )}{c}+\frac {2 \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )}{c}-\frac {2 \pi \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{c}+\frac {\pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )}{c}-\frac {2 i \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{c}}{4 c^2}-\frac {\frac {i \pi \text {ArcSin}(c x)}{2 c}-\frac {i \text {ArcSin}(c x)^2}{2 c}+\frac {2 \pi \log \left (1+e^{-i \text {ArcSin}(c x)}\right )}{c}+\frac {\pi \log \left (1-i e^{i \text {ArcSin}(c x)}\right )}{c}+\frac {2 \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )}{c}-\frac {2 \pi \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )}{c}-\frac {\pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )}{c}-\frac {2 i \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{c}}{4 c^2}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcSin[c*x]))/(d - c^2*d*x^2)^2,x]

[Out]

-1/2*(a*x)/(c^2*d^2*(-1 + c^2*x^2)) + (a*Log[1 - c*x])/(4*c^3*d^2) - (a*Log[1 + c*x])/(4*c^3*d^2) + (b*((Sqrt[
1 - c^2*x^2] - ArcSin[c*x])/(4*c^3*(-1 + c*x)) - (Sqrt[1 - c^2*x^2] + ArcSin[c*x])/(4*c^2*(c + c^2*x)) + ((((3
*I)/2)*Pi*ArcSin[c*x])/c - ((I/2)*ArcSin[c*x]^2)/c + (2*Pi*Log[1 + E^((-I)*ArcSin[c*x])])/c - (Pi*Log[1 + I*E^
(I*ArcSin[c*x])])/c + (2*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])])/c - (2*Pi*Log[Cos[ArcSin[c*x]/2]])/c + (Pi*
Log[-Cos[(Pi + 2*ArcSin[c*x])/4]])/c - ((2*I)*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/c)/(4*c^2) - (((I/2)*Pi*ArcS
in[c*x])/c - ((I/2)*ArcSin[c*x]^2)/c + (2*Pi*Log[1 + E^((-I)*ArcSin[c*x])])/c + (Pi*Log[1 - I*E^(I*ArcSin[c*x]
)])/c + (2*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])])/c - (2*Pi*Log[Cos[ArcSin[c*x]/2]])/c - (Pi*Log[Sin[(Pi +
2*ArcSin[c*x])/4]])/c - ((2*I)*PolyLog[2, I*E^(I*ArcSin[c*x])])/c)/(4*c^2)))/d^2

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Maple [A]
time = 0.19, size = 238, normalized size = 1.65

method	result	size
derivativedivides	\(\frac {-\frac {a}{4 d^{2} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}+\frac {a \ln \left (c x -1\right )}{4 d^{2}}-\frac {b \arcsin \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}}{c^{3}}\)	\(238\)
default	\(\frac {-\frac {a}{4 d^{2} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}+\frac {a \ln \left (c x -1\right )}{4 d^{2}}-\frac {b \arcsin \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}}{c^{3}}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/c^3*(-1/4*a/d^2/(c*x+1)-1/4*a/d^2*ln(c*x+1)-1/4*a/d^2/(c*x-1)+1/4*a/d^2*ln(c*x-1)-1/2*b/d^2/(c^2*x^2-1)*arcs
in(c*x)*c*x+1/2*b/d^2/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)+1/2*b/d^2*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-
1/2*b/d^2*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/2*I*b/d^2*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+1/2
*I*b/d^2*dilog(1-I*(I*c*x+(-c^2*x^2+1)^(1/2))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

-1/4*a*(2*x/(c^4*d^2*x^2 - c^2*d^2) + log(c*x + 1)/(c^3*d^2) - log(c*x - 1)/(c^3*d^2)) - 1/4*(2*c*x*arctan2(c*
x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (c^2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - (c^
2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) + 4*(c^5*d^2*x^2 - c^3*d^2)*integrate(1/4*
(2*c*x + (c^2*x^2 - 1)*log(c*x + 1) - (c^2*x^2 - 1)*log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^6*d^2*x^4 -
2*c^4*d^2*x^2 + c^2*d^2), x))*b/(c^5*d^2*x^2 - c^3*d^2)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*x^2*arcsin(c*x) + a*x^2)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a x^{2}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac {b x^{2} \operatorname {asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asin(c*x))/(-c**2*d*x**2+d)**2,x)

[Out]

(Integral(a*x**2/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b*x**2*asin(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1),
x))/d**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*arcsin(c*x) + a)*x^2/(c^2*d*x^2 - d)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*asin(c*x)))/(d - c^2*d*x^2)^2,x)

[Out]

int((x^2*(a + b*asin(c*x)))/(d - c^2*d*x^2)^2, x)

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